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  • AD is a diameter of a circle and AB is a chord. If AD equal to 34 cm, AB equal to 30 cm, the distance of AB from the center of the circle is: (A) 17 cm(B) 15 cm(C) 4 cm(D) 8 cm

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the center of the circle is:

(A) 17 cm

(B) 15 cm

(C) 4 cm

(D) 8 cm

Answers (1)

(D) 8 cm

Hint:

Solution:

Given AD = 34 cm and AB = 30 cm

In figure, draw OL \perp AB

AL = LB = \frac{1}{2} AB = 15 cm

 In Right angled \triangleOLA

OA2 = OL2 + AL2

(By Pythagoras Theorem)

(17)2 = OL2 + (15)2

OL2 = 289 – 225 = 64

OL = 8 cm

(Taking positive square root, because length is always positive)

Hence the distance of the chord from the centre is 8 cm.

Therefore option (D) is correct.

 

Posted by

infoexpert24

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